The Fibonacci numbers are the numbers in the following integer sequence.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

    Fn = Fn-1 + Fn-2

with seed values

F0 = 0 and F1 = 1.

Write a function int fib(int n) that returns F_n. For example, if n = 0, then fib() should return 0. If n = 1, then it should return 1. For n > 1, it should return F_n-1 + F_n-2

For n = 9
Output:34

Following are different methods to get the nth Fibonacci number.

Method 1 ( Space Optimized )

We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibonacci number in series.

// Fibonacci Series using Space Optimized Method
#include<stdio.h>
int fib(int n)
{
  int a = 0, b = 1, c, i;
  if( n == 0)
    return a;
  for (i = 2; i <= n; i++)
  {
     c = a + b;
     a = b;
     b = c;
  }
  return b;
}
 
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Time Complexity: O(n)
Extra Space: O(1)

Method 2 (O(Log n) Time)
Below is one more interesting recurrence formula that can be used to find n’th Fibonacci Number in O(Log n) time.

If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)

If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)

How does this formula work?
The formula can be derived from above matrix equation.

Taking determinant on both sides, we get
(-1)ⁿ = F_n+1F_n-1 – F_n²
Moreover, since AⁿA^m = A^n+m for any square matrix A, the following identities can be derived (they are obtained form two different coefficients of the matrix product)

F_mF_n + F_m-1F_n-1 = F_m+n-1

By putting n = n+1,

F_mF_n+1 + F_m-1F_n = F_m+n

Putting m = n

F_2n-1 = F_n² + F_n-1²

F_2n = (F_n-1 + F_n+1)F_n = (2F_n-1 + F_n)F_n (Source: Wiki)

To get the formula to be proved, we simply need to do following
If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2

Below is C++ implementation of above idea.

// C++ Program to find n'th fibonacci Number in
// with O(Log n) arithmatic operations
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1000;
 
// Create an array for memoization
int f[MAX] = {0};
 
// Returns n'th fuibonacci number using table f[]
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n])
        return f[n];
 
    int k = (n & 1)? (n+1)/2 : n/2;
 
    // Applyting above formula [Note value n&1 is 1
    // if n is odd, else 0.
    f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
           : (2*fib(k-1) + fib(k))*fib(k);
 
    return f[n];
}
 
/* Driver program to test above function */
int main()
{
    int n = 9;
    printf("%d ", fib(n));
    return 0;
}

Output :

34

Time complexity of this solution is O(Log n) as we divide the problem to half in every recursive call.